Đáp án:
\(\begin{array}{l}
1,\\
a,\\
x \ge \dfrac{1}{3}\\
b,\\
x \le \dfrac{5}{3}\\
c,\\
2 \le x \le 4\\
d,\\
x > 2\\
e,\\
x > 2\\
f,\\
- \dfrac{2}{7} < x \le 3\\
g,\\
- 2 \le x < \dfrac{7}{2}\\
h,\\
\left\{ \begin{array}{l}
x \ge - 1\\
x \ne 0
\end{array} \right.\\
i,\\
x > 1\\
2,\\
a,\\
2 > \sqrt 3 \\
6 < \sqrt {41} \\
7 > \sqrt {47} \\
b,\\
2\sqrt {31} > 10\\
- 3\sqrt {11} > - 12\\
3,\\
a,\\
{\left( {\sqrt 3 - 1} \right)^2}\\
b,\\
{\left( {2 + \sqrt 3 } \right)^2}\\
c,\\
{\left( {2\sqrt 3 - 1} \right)^2}\\
4,\\
A = 0\\
B = - 11\sqrt 5 \\
C = 2\sqrt 2 \\
D = - 2\sqrt 3 \\
5,\\
a,\\
A = 6 - 2\sqrt a \\
b,\\
B = 0
\end{array}\)
Giải thích các bước giải:
Bài 1:
Các biểu thức đã cho có nghĩa khi:
\(\begin{array}{l}
a,\\
3x - 1 \ge 0 \Leftrightarrow 3x \ge 1 \Leftrightarrow x \ge \dfrac{1}{3}\\
b,\\
5 - 3x \ge 0 \Leftrightarrow 5 \ge 3x \Leftrightarrow 3x \le 5 \Leftrightarrow x \le \dfrac{5}{3}\\
c,\\
\left\{ \begin{array}{l}
x - 2 \ge 0\\
4 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
4 \ge x
\end{array} \right. \Leftrightarrow 2 \le x \le 4\\
d,\\
\left\{ \begin{array}{l}
x - 2 \ge 0\\
{x^2} - 4 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
{x^2} \ne 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \ne 2\\
x \ne - 2
\end{array} \right. \Leftrightarrow x > 2\\
e,\\
\left\{ \begin{array}{l}
7x - 14 \ge 0\\
\sqrt {7x - 14} \ne 0
\end{array} \right. \Leftrightarrow 7x - 14 > 0 \Leftrightarrow 7x > 14 \Leftrightarrow x > 2\\
f,\\
\left\{ \begin{array}{l}
3 - x \ge 0\\
7x + 2 \ge 0\\
\sqrt {7x + 2} \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3 - x \ge 0\\
7x + 2 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 3\\
x > - \dfrac{2}{7}
\end{array} \right. \Leftrightarrow - \dfrac{2}{7} < x \le 3\\
g,\\
\left\{ \begin{array}{l}
\dfrac{{x + 2}}{{7 - 2x}} \ge 0\\
7 - 2x \ne 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 \ge 0\\
7 - 2x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 \le 0\\
7 - 2x < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 2\\
7 > 2x
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - 2\\
7 < 2x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 2\\
x < \dfrac{7}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - 2\\
x > \dfrac{7}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow - 2 \le x < \dfrac{7}{2}\\
h,\\
\left\{ \begin{array}{l}
x + 1 \ge 0\\
\sqrt {x + 1} - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
\sqrt {x + 1} \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x + 1 \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \ne 0
\end{array} \right.\\
i,\\
\left\{ \begin{array}{l}
\dfrac{3}{{x - 1}} \ge 0\\
x - 1 \ne 0
\end{array} \right. \Leftrightarrow x - 1 > 0 \Leftrightarrow x > 1\\
2,\\
a,\\
2 = \sqrt {{2^2}} = \sqrt 4 > \sqrt 3 \\
\Rightarrow 2 > \sqrt 3 \\
6 = \sqrt {{6^2}} = \sqrt {36} < \sqrt {41} \\
\Rightarrow 6 < \sqrt {41} \\
7 = \sqrt {{7^2}} = \sqrt {49} > \sqrt {47} \\
\Rightarrow 7 > \sqrt {47} \\
b,\\
2\sqrt {31} > 2\sqrt {25} = 2.\sqrt {{5^2}} = 2.5 = 10\\
\Rightarrow 2\sqrt {31} > 10\\
3\sqrt {11} < 3\sqrt {16} = 3.\sqrt {{4^2}} = 3.4 = 12\\
\Rightarrow 3\sqrt {11} < 12\\
\Rightarrow - 3\sqrt {11} > - 12\\
3,\\
a,\\
4 - 2\sqrt 3 = 3 - 2\sqrt 3 + 1 = {\sqrt 3 ^2} - 2.\sqrt 3 .1 + {1^2} = {\left( {\sqrt 3 - 1} \right)^2}\\
b,\\
7 + 4\sqrt 3 = 4 + 4\sqrt 3 + 3 = {2^2} + 2.2.\sqrt 3 + {\sqrt 3 ^2} = {\left( {2 + \sqrt 3 } \right)^2}\\
c,\\
13 - 4\sqrt 3 = 12 - 4\sqrt 3 + 1 = {\left( {2\sqrt 3 } \right)^2} - 2.2\sqrt 3 .1 + {1^2} = {\left( {2\sqrt 3 - 1} \right)^2}\\
4,\\
A = \sqrt 8 - 5\sqrt {32} + 3\sqrt {72} \\
= \sqrt {4.2} - 5.\sqrt {16.2} + 3\sqrt {36.2} \\
= \sqrt {{2^2}.2} - 5\sqrt {{4^2}.2} + 3.\sqrt {{6^2}.2} \\
= 2\sqrt 2 - 5.4\sqrt 2 + 3.6\sqrt 2 \\
= 2\sqrt 2 - 20\sqrt 2 + 18\sqrt 2 \\
= 0\\
B = \sqrt {20} - 2\sqrt {45} - 3\sqrt {80} + \sqrt {125} \\
= \sqrt {4.5} - 2\sqrt {9.5} - 3\sqrt {16.5} + \sqrt {25.5} \\
= \sqrt {{2^2}.5} - 2\sqrt {{3^2}.5} - 3.\sqrt {{4^2}.5} + \sqrt {{5^2}.5} \\
= 2\sqrt 5 - 2.3\sqrt 5 - 3.4\sqrt 5 + 5\sqrt 5 \\
= 2\sqrt 5 - 6\sqrt 5 - 12\sqrt 5 + 5\sqrt 5 \\
= - 11\sqrt 5 \\
C = \sqrt {3 + 2\sqrt 2 } + \sqrt {3 - 2\sqrt 2 } \\
= \sqrt {2 + 2\sqrt 2 + 1} + \sqrt {2 - 2\sqrt 2 + 1} \\
= \sqrt {{{\sqrt 2 }^2} + 2.\sqrt 2 .1 + {1^2}} + \sqrt {{{\sqrt 2 }^2} - 2.\sqrt 2 .1 + {1^2}} \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
= \left| {\sqrt 2 + 1} \right| + \left| {\sqrt 2 - 1} \right|\\
= \sqrt 2 + 1 + \sqrt 2 - 1\\
= 2\sqrt 2 \\
D = \sqrt {8 - 2\sqrt {15} } - \sqrt {8 + 2\sqrt {15} } \\
= \sqrt {5 - 2\sqrt {15} + 3} - \sqrt {5 + 2\sqrt {15} + 3} \\
= \sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .\sqrt 3 + {{\sqrt 3 }^2}} - \sqrt {{{\sqrt 5 }^2} + 2.\sqrt 5 .\sqrt 3 + {{\sqrt 3 }^2}} \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}} \\
= \left| {\sqrt 5 - \sqrt 3 } \right| - \left| {\sqrt 5 + \sqrt 3 } \right|\\
= \left( {\sqrt 5 - \sqrt 3 } \right) - \left( {\sqrt 5 + \sqrt 3 } \right)\\
= \sqrt 5 - \sqrt 3 - \sqrt 5 - \sqrt 3 \\
= - 2\sqrt 3 \\
5,\\
a,\\
A = \dfrac{{9 - a}}{{\sqrt a + 3}} - \dfrac{{9 - 6\sqrt a + a}}{{\sqrt a - 3}}\\
= \dfrac{{{3^3} - {{\sqrt a }^2}}}{{\sqrt a + 3}} - \dfrac{{{{\sqrt a }^2} - 2.\sqrt a .3 + {3^2}}}{{\sqrt a - 3}}\\
= \dfrac{{\left( {3 - \sqrt a } \right)\left( {3 + \sqrt a } \right)}}{{\sqrt a + 3}} - \dfrac{{{{\left( {\sqrt a - 3} \right)}^2}}}{{\sqrt a - 3}}\\
= \left( {3 - \sqrt a } \right) - \left( {\sqrt a - 3} \right)\\
= 3 - \sqrt a - \sqrt a + 3\\
= 6 - 2\sqrt a \\
b,\\
B = \dfrac{{a + b - 2\sqrt {ab} }}{{\sqrt a - \sqrt b }} - \dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{{{\sqrt a }^2} - 2.\sqrt a .\sqrt b + {{\sqrt b }^2}}}{{\sqrt a - \sqrt b }} - \dfrac{{{{\sqrt a }^2} - {{\sqrt b }^2}}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2}}}{{\sqrt a - \sqrt b }} - \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a + \sqrt b }}\\
= \left( {\sqrt a - \sqrt b } \right) - \left( {\sqrt a - \sqrt b } \right)\\
= 0
\end{array}\)
